H = $${1 \over 2}$$gt²

$${{9.8 \times 2} \over {9.8}}$$ = t²

t = $$\sqrt 2 $$ sec

$$\Delta$$t : time interval between drops

h = $${1 \over 2}$$g($$\sqrt 2 $$ $$-$$ 2$$\Delta$$t)²

$$\Delta$$t = $${1 \over {\sqrt 2 }}$$

h = $${1 \over 2}$$g$${\left( {\sqrt 2 - {1 \over {\sqrt 2 }}} \right)^2} = {1 \over 2} \times 9.8 \times {1 \over 2} = {{9.8} \over 4} = 2.45$$ m

H $$-$$ h = 9.8 $$-$$ 2.45

= 7.35 m

Relative velocity of bomb w.r.t. observer in plane = 0.

Bomb will fall down vertically. So, it will move in straight line w.r.t. observer.

$$u = \sqrt {2gh} $$

Now,

$$S = {h \over 3}$$

a = $$-$$g

$$S = ut + {1 \over 2}a{t^2}$$

$${h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}$$

$${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$$

From quadratic equation

$${t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}$$

$${{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$

$$V = \alpha t + \beta {t^2}$$

$${{ds} \over {dt}} = \alpha t + \beta {t^2}$$

$$\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} } $$

$${S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2$$

As particle is not changing direction

So distance = displacement

Distance = $$\left[ {{{\alpha [4 - 1]} \over 2} + {{\beta [8 - 1]} \over 3}} \right]$$

$$ = {{3\alpha } \over 2} + {{7\beta } \over 3}$$